= Solution
The <kinship coefficient> $\phi_{XY}$ is the <probability> that a randomly selected <allele> copy from individual $X$ and an independently selected copy from individual $Y$ at the same <genetic locus> are <identical by descent>. The sampling is of gene copies, not of whole <genotypes>. For a noninbred individual $X$, sampling its two copies with replacement gives $\phi_{XX}=1/2$; in general $\phi_{XX}=(1+F_X)/2$, where $F_X$ is its <inbreeding coefficient>.
For two <full siblings> in an <outbred pedigree>, the two sampled copies are both paternal with <probability> $1/4$, both maternal with <probability> $1/4$, and of different parental origin with <probability> $1/2$. Given that both are paternal, their independent <Mendelian segregation> transmissions select the same paternal ancestral copy with <probability> $1/2$; likewise for both maternal. The different-parent case cannot be <identical by descent> when the parents are unrelated. Therefore
$$
\boxed{\phi_{\mathrm{sib}}=\frac14\cdot\frac12+\frac14\cdot\frac12+\frac12\cdot0=\frac14.}
$$
Equivalently, conditional on an IBD-sharing count $J$, the chance that the two random copies match is $J/4$, so $\phi_{\mathrm{sib}}=E[J]/4=1/4$. The expected shared proportion $E[J/2]=1/2$ and the usual additive relationship coefficient $2\phi=1/2$ are twice this <kinship coefficient>; confusing these conventions would double the subsequent <inbreeding coefficient>.
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